# 双摆

## 分析以及詮釋

{\displaystyle {\begin{aligned}x_{1}&={\frac {l}{2}}\sin \theta _{1}\\y_{1}&=-{\frac {l}{2}}\cos \theta _{1}\end{aligned}}}

{\displaystyle {\begin{aligned}x_{2}&=l\left(\sin \theta _{1}+{\tfrac {1}{2}}\sin \theta _{2}\right)\\y_{2}&=-l\left(\cos \theta _{1}+{\tfrac {1}{2}}\cos \theta _{2}\right)\end{aligned}}}

### 拉格朗日力學

{\displaystyle {\begin{aligned}L&={\text{kinetic energy}}-{\text{potential energy}}\\&={\tfrac {1}{2}}m\left(v_{1}^{2}+v_{2}^{2}\right)+{\tfrac {1}{2}}I\left({{\dot {\theta }}_{1}}^{2}+{{\dot {\theta }}_{2}}^{2}\right)-mg\left(y_{1}+y_{2}\right)\\&={\tfrac {1}{2}}m\left({{\dot {x}}_{1}}^{2}+{{\dot {y}}_{1}}^{2}+{{\dot {x}}_{2}}^{2}+{{\dot {y}}_{2}}^{2}\right)+{\tfrac {1}{2}}I\left({{\dot {\theta }}_{1}}^{2}+{{\dot {\theta }}_{2}}^{2}\right)-mg\left(y_{1}+y_{2}\right)\end{aligned}}}

${\displaystyle L={\tfrac {1}{6}}ml^{2}\left({{\dot {\theta }}_{2}}^{2}+4{{\dot {\theta }}_{1}}^{2}+3{{\dot {\theta }}_{1}}{{\dot {\theta }}_{2}}\cos(\theta _{1}-\theta _{2})\right)+{\tfrac {1}{2}}mgl\left(3\cos \theta _{1}+\cos \theta _{2}\right).}$

{\displaystyle {\begin{aligned}p_{\theta _{1}}&={\frac {\partial L}{\partial {{\dot {\theta }}_{1}}}}={\tfrac {1}{6}}ml^{2}\left(8{{\dot {\theta }}_{1}}+3{{\dot {\theta }}_{2}}\cos(\theta _{1}-\theta _{2})\right)\\p_{\theta _{2}}&={\frac {\partial L}{\partial {{\dot {\theta }}_{2}}}}={\tfrac {1}{6}}ml^{2}\left(2{{\dot {\theta }}_{2}}+3{{\dot {\theta }}_{1}}\cos(\theta _{1}-\theta _{2})\right).\end{aligned}}}

{\displaystyle {\begin{aligned}{{\dot {\theta }}_{1}}&={\frac {6}{ml^{2}}}{\frac {2p_{\theta _{1}}-3\cos(\theta _{1}-\theta _{2})p_{\theta _{2}}}{16-9\cos ^{2}(\theta _{1}-\theta _{2})}}\\{{\dot {\theta }}_{2}}&={\frac {6}{ml^{2}}}{\frac {8p_{\theta _{2}}-3\cos(\theta _{1}-\theta _{2})p_{\theta _{1}}}{16-9\cos ^{2}(\theta _{1}-\theta _{2})}}.\end{aligned}}}

{\displaystyle {\begin{aligned}{{\dot {p}}_{\theta _{1}}}&={\frac {\partial L}{\partial \theta _{1}}}=-{\tfrac {1}{2}}ml^{2}\left({{\dot {\theta }}_{1}}{{\dot {\theta }}_{2}}\sin(\theta _{1}-\theta _{2})+3{\frac {g}{l}}\sin \theta _{1}\right)\\{{\dot {p}}_{\theta _{2}}}&={\frac {\partial L}{\partial \theta _{2}}}=-{\tfrac {1}{2}}ml^{2}\left(-{{\dot {\theta }}_{1}}{{\dot {\theta }}_{2}}\sin(\theta _{1}-\theta _{2})+{\frac {g}{l}}\sin \theta _{2}\right).\end{aligned}}}

## 混沌運動

• 10lg（綠色）
• 100lg（紅色）
• 1000lg（紫色）
• 10000lg（蓝色）

${\displaystyle 3\cos \theta _{1}+\cos \theta _{2}=2.}$

${\displaystyle 3\cos \theta _{1}+\cos \theta _{2}>2,}$

## 参考資料

1. ^ 『機械工学辞典』 日本機械学会、丸善、2007年1月20日、第2版、966頁。ISBN 978-4-88898-083-8
2. ^ Levien RB and Tan SM. Double Pendulum: An experiment in chaos.American Journal of Physics 1993; 61 (11): 1038
3. ^ Alex Small, Sample Final Project: One Signature of Chaos in the Double Pendulum[失效連結], (2013). A report produced as an example for students. Includes a derivation of the equations of motion, and a comparison between the double pendulum with 2 point masses and the double pendulum with 2 rods.